[EM] Election-Methods Digest, Vol 213, Issue 44
Richard Lung
voting at ukscientists.com
Wed May 4 11:31:44 PDT 2022
Steve,
Thanks for reply. A basic objection to graded classes is that it takes ordered numbers to complete the transition from the classificatory scale to the ordinal scale, in S S Stevens widely accepted scales of measurement, in the sciences: nominal, ordinal, interval, ratio scales. The vote should be an ordered number vote. (Its classificatory scale is one person one vote.) The count is an interval scale and a ratio scale count.
The ordinal scale founds, in the sciences, sufficiently advanced, an orderly transition to the two most powerful scales.
The stripped-down version of FAB STV, is simple enough to enable a Binomial STV hand count, and can elect single members, as mentioned, using a rational exclusion count, as well as a rational election count.
Regards,
Richard Lung.
On 3 May 2022, at 9:30 pm, steve bosworth <stevebosworth at hotmail.com> wrote:
Re: Re: [EM] Election-Methods Digest, Vol 213, Issue 44
Richard,
Sorry, with you initially starting out with President Macron, I mistakenly responded to your argument for FAB STV as if you were proposing it as a single-winner method. However, your most recent post copied below now makes it clear that you instead see it as a superior multi-winner method. Thank you.
In this light, I would like to ask you to compare FAB STV with the multi-winner method my co-authors and I call evaluative proportional representation (EPR – https://www.jpolrisk.com/legislatures-elected-by-evaluative-proportional-representation-epr-an-algorithm-v3/). We see EPR as an improved version of ordinary STV that follows MJ by inviting citizens to rank candidates more informatively instead by grading them – grading their suitability for office as either Excellent, Very Good, Good, Acceptable, Poor, or Reject. We have also updated theabove 2020 article in a Paper that I would be happy to email to any EM reader upon request (stevebosworth at hotmail.com).
I see EPR as also fully satisfying your desire to include the judgments of the whole electorate in the count to the fullest extent possible to avoid *any … minimally democratic ... binary choice* as you put it.
With regard to FAB STV, I am happy to assume that it provides this benefit much more than plurality and ordinary STV does. At the same time, please tell me how the results of an FAB STV election at-large of a seven-member city council would compare with the following results that an EPR election guarantees: every citizens’ ballot equally adds to the voting power (weighted vote) in the council of the elected candidate each sees as likely to represent their hopes and concerns most accurately. This winner will have received either this citizen’s highest grade, remaining highest grade, or proxy vote. Consequently, every citizen’s vote cast is equally represented in the council quantitatively. Exactly how EPR offers these democratic benefits is described in the above mentioned available Paper.
You (Richard) freely warn readers that most of them may not be able fully to understand the complex steps by which an FAB STV election would be counted. In contrast, my co-authors and I believe that anyone who can count, add, and subtract whole numbers will be able to understand exactly how an EPR election is counted.
I look forward to our dialogues.
Steve
From: Richard Lung <voting at ukscientists.com>
Sent: Monday, May 2, 2022 11:16 AM
To: steve bosworth <stevebosworth at hotmail.com>
Subject: Re: [EM] Election-Methods Digest, Vol 213, Issue 44
Steve,
Single winner elections are the least democratic; they give too much power to one person. A problem of peace is how to get politicians away from the fixation with the monarchic principle. Lord Hailsham, in The Dilemma of Democracy, said Britain is an "elective dictatorship". This serves as a definition of a tyranny, in the classic Greek sense.
FAB STV is a system designed for a democratic minimum representation of a 4 or 5 member district. I was asked for a hand count version, which is much simpler, and I just call Binomial STV. It involves an election count divided by an exclusion count of all candidates keep values. This does away with haphazardly eliminating candidates, when the surplus transfers run out.
All the preference information is counted, including abstentions. This measures one whole dimension of choice, which, as far as I know, is unique in election methods.
Regards,
Richard Lung.
>>>>>>>>>>>>>>>>>>>>>>>>>>>
From: Richard Lung <voting at ukscientists.com>
Sent: Sunday, May 1, 2022 4:45 AM
To: steve bosworth <stevebosworth at hotmail.com>
Cc: election-methods at lists.electorama.com <election-methods at lists.electorama.com>
Subject: Re: [EM] Election-Methods Digest, Vol 213, Issue 44
Richard,
I agree with your desire to have as many citizens as possible in addition to a majority to help determine the single winner – to include the judgments of the whole electorate in the count to the fullest extent possible to avoid *any … minimally democratic ... binary choice*.Also, I am happy to assume that FAB STV provides this benefit much more than plurality and ordinary STV does.
At the same time, do you agree that MJ also does this to some extent in that the median grade received by each candidate is equally determined by all the grades given to all the candidates by all the citizens? Also, am I correct in assuming that you agree that MJ’s counting method for determining the median-grade winner is much easier for people to understand than all the calculations needed to determine the FAB STV winner? My summary of this MJ count is as follows:
The MJ winner is the one who receives an absolute majority of all the grades equal to, or higher than, the highest median grade given to any candidate. This median grade can be found as follows:
1. Place all the grades given to each candidate, high to low, left to right in a row, with the name of each candidate on the left of each row.
2. The median grade for each candidate is in the middle of each row. Specifically, the middle grade for an odd number of voters, or the grade on the right in the middle for an even number of voters.
3. If more than one candidate has the same highest median grade, the MJ winner is discovered by removing, one by one, any grades equal in value to the current highest median grade from each tied candidate’s total until only one of the tied candidates has the highest remaining median grade.
Balinski sees the six grades Excellent to Reject as easy to use. They are intellectual creations which can be *given very careful definitions* and that become more meaningful and precise with the passage of time and use. They have *no meaning other than what is ascribed to them by their users* (pp. 166-169). Such intellectual *creations must always remain a work in progress* (pp.168-70).
However, where we currently seem to disagree is over the validity of your claim that * Ranked choice ... uses a more powerful scale of measurement* when compared with MJ. Presumably, regardless of the voting method an election uses (plurality, IRV (simply ranking the candidates), approval, range, or MJ), each citizen when voting is trying to express their judgments about the degree one or more of the candidates is suitable for the office according to their own perspective. FAB STV is a variety of IRV and so your claim is that rankings provide a more “powerful measure” of these judgments than does MJ’s grades – that rankings are more meaningfully and informatively expressive than the grades MJ citizens use on their ballots. If so, please explain because I see the opposite as true. For example, when we see the 1st preference candidate on someone’s ballot, we do not know whether this voter judged this candidate to be Excellent, Very Good, Good, or Acceptable or least bad. In contrast, some preferences can be inferred from a ballot that lists grades for candidates, e.g. each Excellent candidate is preferred over each Very Good candidate, etc. This is why I see grades as more meaningfully informative about the subjective judgments citizens have made about the candidates than the preferences they would be required to indicate under FAB STV.
If the above reasoning is sound, MJ may be no better than FAB STV in that both use all the votes cast to elect the single winner. However, MJ would still be democratically superior because more citizens would understand exactly how all the votes are counted by MJ, plus MJ also allows every citizen more meaningfully to express themselves.
What do you think?
>>>>>>>>>>>>>1st
Macron says he will represent all France, not just those who voted for him. They say that in the British single member system, too. Only those who actually represent one can represent one. Patronage is not democracy. Single majorities are minimally democratic, leaving out half the voters.
And binary choice, as in the x-vote, is a minimal choice. Approve/disapprove is not much more choice. And more categories of choice not much better still. Ranked choice or number order choice uses a more powerful scale of measurement, the ordinal scale vote, instead of the categorical scale. This makes possible more precise measurement. For instance, more precise averages than the median. (My system of FAB STV uses four rational averages.) "Science is measurement."
Regards,
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
To: steve bosworth <stevebosworth at hotmail.com>
Subject: Re: [EM] Election-Methods Digest, Vol 213, Issue 44
Steve,
Single winner elections are the least democratic; they give too much power to one person. A problem of peace is how to get politicians away from the fixation with the monarchic principle. Lord Hailsham, in The Dilemma of Democracy, said Britain is an "elective dictatorship". This serves as a definition of a tyranny, in the classic Greek sense.
FAB STV is a system designed for a democratic minimum representation of a 4 or 5 member district. I was asked for a hand count version, which is much simpler, and I just call Binomial STV. It involves an election count divided by an exclusion count of all candidates keep values. This does away with haphazardly eliminating candidates, when the surplus transfers run out.
All the preference information is counted, including abstentions. This measures one whole dimension of choice, which, as far as I know, is unique in election methods.
Regards,
Richard Lung.
On 2 May 2022, at 7:44 am, steve bosworth <stevebosworth at hotmail.com> wrote:
From: Richard Lung <voting at ukscientists.com>
Sent: Sunday, May 1, 2022 4:45 AM
To: steve bosworth <stevebosworth at hotmail.com>
Cc: election-methods at lists.electorama.com <election-methods at lists.electorama.com>
Subject: Re: [EM] Election-Methods Digest, Vol 213, Issue 44
Richard,
I agree with your desire to have as many citizens as possible in addition to a majority to help determine the single winner – to include the judgments of the whole electorate in the count to the fullest extent possible to avoid *any … minimally democratic ... binary choice*.Also, I am happy to assume that FAB STV provides this benefit much more than plurality and ordinary STV does.
At the same time, do you agree that MJ also does this to some extent in that the median grade received by each candidate is equally determined by all the grades given to all the candidates by all the citizens? Also, am I correct in assuming that you agree that MJ’s counting method for determining the median-grade winner is much easier for people to understand than all the calculations needed to determine the FAB STV winner? My summary of this MJ count is as follows:
The MJ winner is the one who receives an absolute majority of all the grades equal to, or higher than, the highest median grade given to any candidate. This median grade can be found as follows:
1. Place all the grades given to each candidate, high to low, left to right in a row, with the name of each candidate on the left of each row.
2. The median grade for each candidate is in the middle of each row. Specifically, the middle grade for an odd number of voters, or the grade on the right in the middle for an even number of voters.
3. If more than one candidate has the same highest median grade, the MJ winner is discovered by removing, one by one, any grades equal in value to the current highest median grade from each tied candidate’s total until only one of the tied candidates has the highest remaining median grade.
Balinski sees the six grades Excellent to Reject as easy to use. They are intellectual creations which can be *given very careful definitions* and that become more meaningful and precise with the passage of time and use. They have *no meaning other than what is ascribed to them by their users* (pp. 166-169). Such intellectual *creations must always remain a work in progress* (pp.168-70).
However, where we currently seem to disagree is over the validity of your claim that * Ranked choice ... uses a more powerful scale of measurement* when compared with MJ. Presumably, regardless of the voting method an election uses (plurality, IRV (simply ranking the candidates), approval, range, or MJ), each citizen when voting is trying to express their judgments about the degree one or more of the candidates is suitable for the office according to their own perspective. FAB STV is a variety of IRV and so your claim is that rankings provide a more “powerful measure” of these judgments than does MJ’s grades – that rankings are more meaningfully and informatively expressive than the grades MJ citizens use on their ballots. If so, please explain because I see the opposite as true. For example, when we see the 1st preference candidate on someone’s ballot, we do not know whether this voter judged this candidate to be Excellent, Very Good, Good, or Acceptable or least bad. In contrast, some preferences can be inferred from a ballot that lists grades for candidates, e.g. each Excellent candidate is preferred over each Very Good candidate, etc. This is why I see grades as more meaningfully informative about the subjective judgments citizens have made about the candidates than the preferences they would be required to indicate under FAB STV.
If the above reasoning is sound, MJ may be no better than FAB STV in that both use all the votes cast to elect the single winner. However, MJ would still be democratically superior because more citizens would understand exactly how all the votes are counted by MJ, plus MJ also allows every citizen more meaningfully to express themselves.
What do you think?
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Steve,
Macron says he will represent all France, not just those who voted for him. They say that in the British single member system, too. Only those who actually represent one can represent one. Patronage is not democracy. Single majorities are minimally democratic, leaving out half the voters.
And binary choice, as in the x-vote, is a minimal choice. Approve/disapprove is not much more choice. And more categories of choice not much better still. Ranked choice or number order choice uses a more powerful scale of measurement, the ordinal scale vote, instead of the categorical scale. This makes possible more precise measurement. For instance, more precise averages than the median. (My system of FAB STV uses four rational averages.) "Science is measurement."
Regards,
Richard Lung.
On 1 May 2022, at 8:25 am, steve bosworth <stevebosworth at hotmail.com> wrote:
From: Election-Methods <election-methods-bounces at lists.electorama.com> on behalf of election-methods-request at lists.electorama.com <election-methods-request at lists.electorama.com>
Sent: Friday, April 29, 2022 1:02 PM
To: election-methods at lists.electorama.com <election-methods at lists.electorama.com>
Subject: Election-Methods Digest, Vol 213, Issue 44
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Today's Topics:
1. Re: French Presidential Second Ballot. (robert bristow-johnson)
2. Re: Three Category Symmetric MJ (robert bristow-johnson)
----------------------------------------------------------------------
Message: 1
Date: Fri, 29 Apr 2022 02:21:02 -0400 (EDT)
From: robert bristow-johnson <rbj at audioimagination.com>
To: election-methods at lists.electorama.com
Subject: Re: [EM] French Presidential Second Ballot.
Message-ID: <1750062174.1088604.1651213262341 at privateemail.com>
Content-Type: text/plain; charset=UTF-8
> On 04/28/2022 6:23 PM Richard Lung <voting at ukscientists.com> wrote:
>
>
> The French Second Ballot illustrates what's wrong with Condorcet Pairing, in general.
> As Simon Laplace first appreciated, you cannot add first and second preferences, because they are of unequal importance. Less than half the votes, in the first ballot, reveal a much greater level of support than do the extra votes gathered in the Second Ballot, exclusive to merely two candidates.
IMO, the "importance" of a vote that should be equal, is that which is associated with the voter's franchise. However the election works out, if fewer voters like A rather than B than the other way around, if A is elected that means those fewer votes had more juice in them and counted more than those votes from the voters supporting B over A.
It's not about their absolute rank on the ballot but about their relative rank on the ballot and every voter's vote counts equally.
That's why I'm a Condorcet disciple.
r b-j . _ . _ . _ . _ rbj at audioimagination.com
"Imagination is more important than knowledge."
From: Stephen Bosworth <stevebosworth at hotmail.com.
To: EM <election-methods at lists.electorama.com>
Subject: Re: [EM] Three Category Symmetric MJ
Correctly, prompted by the recent French Presidential Election, Richard Lung points out one of the democratic flaws in “Condorcet Pairing.” However, Robert Bristow-Johnson still declares that he is a “Condorcet disciple.”
Especially given that Balinski and Laraki conducted an earlier-voluntary MJ election of a French President, what reason can any contributors to EM give for not seeing MJ as the most democratic method for electing presidents?
Only MJ maximizes the probability that, even when many candidates are running, such a president will be supported by an absolute majority of all the votes cast – having received the largest number of grades at least as high as the highest median grade awarded to any of the candidates? By default, each citizen is counted as having Rejected every candidate’s suitability for office except those they have explicitly graded as either Excellent, Very Good, Good, or Acceptable. Only MJ would seem fully to make each “vote … equal” as Richard appropriately recommends. In contrast to rankings, these grades invite citizens more meaningfully and informatively to express their different judgments of the candidates. In turn, this also allows MJ post-election reports most revealingly to be analyzed so as to inform all citizens about the different intensities and numbers of their fellows who are supporting the many different political agendas in their society.
What do you think?
Steve
------------------------------
Message: 2
Date: Fri, 29 Apr 2022 02:34:51 -0400 (EDT)
From: robert bristow-johnson <rbj at audioimagination.com>
To: EM <election-methods at lists.electorama.com>
Subject: Re: [EM] Three Category Symmetric MJ
Message-ID: <126456545.1088965.1651214091642 at privateemail.com>
Content-Type: text/plain; charset=UTF-8
> On 04/28/2022 11:10 PM Forest Simmons <forest.simmons21 at gmail.com> wrote:
>
>
> Cool how everything is related to Fourier Transforms!
>
>
that was just one app. I mainly used it for autocorrelation (never transforming into the frequency domain) and once or twice used it with an FFT to find a peak's location.
But it's about peaks, doesn't matter what the graph is about.
If y[m] > y[m-1] and y[m] > y[m+1] (m integer)
Then you have a discrete peak at m but an implied peak a little to the left or right of m. I think that the more "precise" location of the peak is
n = m + (1/2)(y[m+1]-y[m-1])/(2y[m]-y[m+1]-y[m-1])
but it assumes a quadratic fit. (But that's usually good enough.)
> El jue., 28 de abr. de 2022 2:02 p. m., robert bristow-johnson <rbj at audioimagination.com> escribi?:
> >
> > that looks a lot like the quadratic interpolation of a peak's locations when you have three discrete points located at
> >
> > (-1, x), (0, 1/2), and (+1, y)
> >
> > where both x <= 1/2 and y <= 1/2 (there's a zero-over-zero problem if there is this equality x = y = 1/2.
> >
> > I use this formula often with auto-correlation (used for musical pitch detection) and sometimes with the magnitude of spectrum that comes from the FFT.
> >
> > interesting to see something familiar.
> >
> > r b-j
> >
> > > On 04/28/2022 1:55 PM Forest Simmons <forest.simmons21 at gmail.com> wrote:
> > >
> > >
> > > contourplot (.5(y-x)/(1-x-y) for x=0... .5, y=0... .5)
> > >
> > >
> > > Paste the above command into Wolfram Alpha to get a contour plot of the Middle region.
> > >
> > >
> > > El mi?., 27 de abr. de 2022 6:10 p. m., Forest Simmons <forest.simmons21 at gmail.com> escribi?:
> > > > A brief summary: The three judgment categories are Good, Middle, and Bad, meaning Satisfactory to Excellent, Mediocre, and Unsuitable (for whatever reasons, including inability to elicit an opinion from the voter, so blank=Bad).
> > > >
> > > > Candidates marked Good on more than half of the ballots are judged to be in the Good category. Candidates marked Bad on more than half of the ballots are judged to be in the Bad category. All other candidates, including those marked Mediocre on more than half of the ballots, are judged to be in the Middle category.
> > > >
> > > > It is desirable for election purposes to establish a finish order that respects these judgment categories ... with all candidates judged Good, ahead of the the other candidates, and all candidates judged Bad behind the other candidates.
> > > >
> > > > Within the Good category, candidate k finishes ahead of candidate j if the number of ballots g(k) on which k is marked Good exceeds the corresponding number g(j) for candidate j. If this rule needs further resolution because j and k are marked good on the same number of ballots, then k is ahead of j in the finish orde if k is marked Bad on fewer ballots than j is so marked.
> > > >
> > > > Similarly, within the Bad category, k finishes ahead of j if k is marked Bad on more ballots than k is, and when j and k are marked Bad on the same number of ballots k is ahead of j if k is marked Good on more ballots than j is.
> > > >
> > > > So far this is all clear and logical. But how to establish a logical, consistent finish order within the Middle category is not so obvious ... until we have a good graphical representation of the three categories and how they fit together.
> > > >
> > > > How they fit together is important because (among other reasons) a small change in the number of ballots can easily bump a borderline candidate from one category into another. If k goes from (barely) Good to Middle, it should enter the Middle category at the upper end of the finish order within the Middle category. Otherwise, the method has no hope of passing any reasonable form of the Participation criterion.
> > > >
> > > > The graphical representation will make this continuity requirement clear. One definition of "topology" is the science of continuity. Without respect for the relevant topology, it isn't possible to have the continuity needed for the Participation compliance alluded to above.
> > > >
> > > > So here's the picture: Let N be the total number of ballots submitted. For each candidate k, let g(k), b(k), and m(k) be the number of ballots on which k is marked Good, Bad, or Mediocre. For graphical purposes let the three dimensional Cartesian coordinates x, y, and z be defined as x=b(k)/N, y=g(k)/N, and z=m(k)/N, so that (no matter the candidate k ... hence suppression of k in the notation) we have the constraint equation x+y+z=1.
> > > >
> > > > The graph of this equation is the convex hull of the three points (1,0,0), (0,1,0), and (0, 0, 1), which are the vertices of an equilateral triangle.
> > > >
> > > > It is convenient to project this graph vertically onto the x,y plane, the planar region R given by the planar inequality x+y<=1. To understand this picture rewrite the constraint equation as x+y=1-z, which reveals z in the role of "slack variable" for the planar inequality.
> > > >
> > > > The intersection of R with the inequality y>50%, is a right triangle representing the set of candidates in the Good category:
> > > > {k | g(k)>50% of N}. The set of candidates tied for y=constant, is a horizontal line segment in the Good triangle. In other words, the "level curves" or "contour lines" of the equations y=c, for .5<c<1-x fill up the Good candidate region with horizontal level curves.
> > > >
> > > > Similarly the vertical segments given by x=c for .5<x<1-y fill up the bad region.
> > > >
> > > > To fill up the Middle region we need a family of line segments that smoothly transition from the vertical segment V forming the leftmost boundary of the Bad region to the horizontal segment H at the lower boundary of the Good region.
> > > >
> > > > The only way to accomplish this is to rotate the segment V clockwise 90 degrees about the point (.5, .5).
> > > >
> > > > At time t let V(t) make a counter clockwise angle with the diagonal y=x of t degrees, t goes from -45 to +45 degrees. Let (x, y) be a point on the segment V(t). Then the angle t between the diagonal y=x and the segment connecting V(t) is 45 degrees minus the angle theta of V(t) with the negative X axis. We have tan(theta) = slope of V(t), which equals (.5 -y)/(.5 -x) since both (x,y) and (.5, .5) are on V(t).
> > > >
> > > > So t = 45 deg - arctan((.5 -y)/(.5 -x))
> > > >
> > > > Taking the tangent of both sides of this equation, and making use of the formula for the tangent of a difference, we get
> > > >
> > > > tan(t) as (tan 45deg - tan theta) divided by (1 +tan 45deg * tan theta).
> > > >
> > > > Since tan 45 deg equals 1, and tan theta equals the slope (.5 -y)/(.5-x), we get
> > > >
> > > > tan(t)= (1 - slope)/(1+slope)
> > > >
> > > > Substituting slope =(.5 -y)/(.5-x), and simplifying algebraically, we get...
> > > >
> > > > tan(t)=(y-x)/[2( 1-x-y)].
> > > >
> > > > Recognizing (1 - x -y) as the slack variable z, we see that tan(t) is just
> > > > (y-x)/(2z). In terms of g, b, and m,
> > > > tan(t)= .5 (g(k)-b(k)/m(k) ....
> > > > which shows where the mysterious formula for the finish order within the Middle category came from in my previous message.
> > > >
> > > > In particular, tan 45deg = 1, which is the same as (.5 -x)/(2(1-x-.5)) given by the formula, and tan(-45 deg) is -1, which is the same as(y-x)/(2z), for any point on V=V(0), where x=.5, and z=1-.5 -y, for any y between 0 and. 5.
> > > >
> > > > It all checks out and fits together at the boundaries of the three regions.
> > > >
> > > > That's it for now!
> > > >
> > > > -Forest
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > El mar., 26 de abr. de 2022 7:20 p. m., Forest Simmons <forest.simmons21 at gmail.com> escribi?:
> > > > > Good Work!
> > > > >
> > > > > I have come to the same conclusion about MJ being much closer to IIA than Range.
> > > > >
> > > > > So I've been trying to improve MJ to make it more symmetrical, satisfy some kind of participation, improve tie breaking all around, etc.
> > > > >
> > > > > I have a good 3-slot version that is as decisive as possible for a reverse symmetry method.
> > > > >
> > > > > I am reminded of our flurry of 3-slot methods from twenty years ago. Back then the EM list was very sure that the best proposals were Condorcet and Approval, but not at all sure which would be most viable. We thought a majoritarian 3-slot method would be plenty simple, easy to count, and have room to distinguish Roosevelt, Stalin, and Hitler. Three slot Bucklin was a popular suggestion with various tie breaking rules, but no version was symmetrical, or any better than (the still unheard of) MJ with three judgment categories.
> > > > >
> > > > > As I have recently come to understand, our lack of design success (i.e. inability to get reverse symmetry into three slot Bucklin) was due to ignorance of the underlying topology.
> > > > >
> > > > > To make a long story short, I will finish this message by cutting straight to the chase, saving the full explanations for tomorrow:
> > > > >
> > > > > Suppose the three categories are Good, Middle, and Bad. Good means a mixture of desirable qualities from competent to excellent. Bad means incompetent or otherwise unsuitable. Middle includes some good qualities but not unalloyed with baser metals ... which is different from "no opinion" the same way that "average" is different from "no basis for a grade." To avoid "dark horse" problems, we will count blank as Bad. (I'm sure some philosopher or lawyer could explain why a candidate able to generate neither appreciable support nor opposition, would be unsuitable.)
> > > > >
> > > > > For each candidate k, let g(k), m(k), & b(k) be the number of ballots on which candidate k is categorized as Good, Middle, or Bad, respectively.
> > > > >
> > > > > If some candidate is judged to be Good on more than half of the ballots, then among the candidates tied for the greatest g(k) value, elect the one categorized as Bad on the fewest ballots.
> > > > >
> > > > > In other words (and symbols) ....
> > > > >
> > > > > elect argmin{b(j)| j is in argmax[g(k)]}.
> > > > >
> > > > > If (in the other extreme) every candidate is judged to be Bad on more than half of the ballots, then from among the candidates tied for the least b(k) value, elect the one categorized as Good on the most ballots. In symbols ...
> > > > >
> > > > > elect argmax{g(j)| j is in argmin[b(k)]}.
> > > > >
> > > > > If neither of the two above cases obtains, then elect from among the candidates in the set argmax[(g(k)-b(k))/m(k)] the candidate j categorized as Bad on the fewest ballots or the one categorized as Good on the most ballots, depending on whether or not g(j) is greater than b(j). In symbols ... elect
> > > > >
> > > > > argmax{T(j)| candidate j is among the tied candidates, i.e. j is a member of argmax[(g(k)-b(k))/m(k)]},
> > > > >
> > > > > where the tie breaker function T to be maximized is given by ...
> > > > >
> > > > > T(j) = min(b(j),g(j))sign(b(j)-g(j))
> > > > >
> > > > > [Here we have made use of the fact that b(j) is minimized when its opposite -b(j) is maximized.]
> > > > >
> > > > > In my next message I will unfold all of these mysteries into plain view!
> > > > >
> > > > > At least now you have the complete 3-slot reverse symmetry compliant MJ recipe safely in the EM cloud.
> > > > >
> > > > > -Forest
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > El mar., 26 de abr. de 2022 2:13 a. m., Kristofer Munsterhjelm <km_elmet at t-online.de> escribi?:
> > > > > > On 26.04.2022 00:51, Forest Simmons wrote:
> > > > > > >
> > > > > > >
> > > > > > > El lun., 25 de abr. de 2022 4:25 a. m., Kristofer Munsterhjelm
> > > > > > > <km_elmet at t-online.de <mailto:km_elmet at t-online.de>> escribi?:
> > > > > > >>
> > > > > > >> So the M-W strategy is: let
> > > > > > >> v_i be the strategic rating we want to find
> > > > > > >> u_i be the public utility of candidate i
> > > > > > >> p_ij be the voter's perceived probability that i and j will
> > > > > > >> be tied.
> > > > > > >>
> > > > > > >
> > > > > > > I could be wrong but I think it should be "tied for winning."
> > > > > >
> > > > > > You're right. I was looking at the paper just now, and it says:
> > > > > >
> > > > > > "For each pair of candidates i and j, the /pivot probability/ p is the
> > > > > > probability (perceived by a voter) of the event that candidates i and j
> > > > > > will be tied for first place in the election."
> > > > > >
> > > > > > I imagine you could refine it a little by letting the p.p. be
> > > > > > parameterized by the vote to submit. E.g. if it's Range voting and i's
> > > > > > score minus j's score is 1, then you could flip the win from i to j by
> > > > > > voting j 10 and i 0. But this would complicate the strategy a lot at
> > > > > > (probably) only very slight benefit.
> > > > > >
> > > > > > > It is interesting that this strategy can actually result in non-solid
> > > > > > > approval coalitions on ballots ... sometimes it requires you to approve
> > > > > > > X while leaving unapproved some candidate Y rated above X on the same
> > > > > > > ballot ... i.e. insincere strategy.
> > > > > > >
> > > > > > > Furthermore, if estimates of both the utilities u_i and u_j, as well as
> > > > > > > of the probabilities p_ij in question were known with a high degree of
> > > > > > > precision, you might get away with those insincere gaps in the approval
> > > > > > > order.?
> > > > > > >
> > > > > > > These facts reflect the fragility (anti-robustness) of the winning tie
> > > > > > > probability based strategy.
> > > > > >
> > > > > > Yes, I think Warren observed something similar: under imperfect
> > > > > > information, the optimal Range/Approval strategy might have you
> > > > > > approving of X and not Y even though you rank Y ahead of X. Under
> > > > > > perfect information, there's always some kind of cutoff where you
> > > > > > approve everybody above it and don't everybody below it.
> > > > > >
> > > > > > > Nevertheless, your result is highly relevant because it shows that on a
> > > > > > > fundamental level there is a meaningful, experimental way of defining
> > > > > > > individual utilities that are just as good as the theoretical utilities
> > > > > > > invoked as a basis for Approval strategy.
> > > > > >
> > > > > > I keep harping on the problem of Range and Approval to fail "de facto
> > > > > > IIA" despite passing it de jure, and I suspect it's related to this. If
> > > > > > we can't standardize a and b, then if the method behaves differently
> > > > > > when given u_i and up_i values, then you can get strange behavior. So
> > > > > > the guidelines about how to vote (mean utility, etc) are just
> > > > > > preprocessing steps that make your ballot expression no longer depend on
> > > > > > what a and b are. Then it's much more honest to attach these guidelines
> > > > > > to the method itself so it does so for the voter, so that voters don't
> > > > > > have to care about what society's a and b values are supposed to be, and
> > > > > > so that the method doesn't get away with sweeping de-facto failures
> > > > > > under the rug.
> > > > > >
> > > > > > At least MJ recognizes this and says "the only way we're going to get
> > > > > > IIA is if we have a and b values that are close enough to commensurable
> > > > > > that the problem doesn't occur". And then the point of using grades
> > > > > > instead of scores, and using order statistics, is to make the whole
> > > > > > process relatively insensitive to what a and b are, so that (hopefully)
> > > > > > a common grade standard can be established.
> > > > > >
> > > > > > > It is equally true for the not as sensitive strategy of approving the
> > > > > > > candidates k with above expectation utilities:?
> > > > > > > u_k >sum P_i u_i,
> > > > > > > based on estimates of (non tie based) winning probabilities P_i, which
> > > > > > > are still sketchy because of rampant misinformation, not to mention
> > > > > > > intentional disinformation.
> > > > > >
> > > > > > Those are zero-info strategies, and indeed, they're also insensitive to
> > > > > > a and b.
> > > > > >
> > > > > > SARVO tries to get around the fragility/chaos problem by averaging over
> > > > > > a lot of vote orders. But it's somewhat of a hack; it's not particularly
> > > > > > elegant, and it fails scale invariance. Perhaps better is finding a
> > > > > > voting equilibrium where the mixed strategy is so that the distribution
> > > > > > of the M-W votes are stable, and then electing the candidate with the
> > > > > > highest expected score. I haven't read the M-W paper in detail, though,
> > > > > > so I don't know if finding this equilibrium is easy.
> > > > > >
> > > > > > (Another possibility, inspired by counterfactual regret minimization, is
> > > > > > to do M-W strategy by every voter, and then once everybdoy has submitted
> > > > > > a vote, pulling one of the voters from the list and having him readjust
> > > > > > his strategic ballot. Keep doing so over a long enough timeline and the
> > > > > > average of scores should converge to an equilibrium.)
> > > > > >
> > > > > > For the zero-info strategies, I tried to figure out what the optimum
> > > > > > zero info strategy is for Lp cumulative voting. I didn't get all the way
> > > > > > there, but this is what I figured:
> > > > > >
> > > > > > Under zero information, p_ij is equal for all pairs, and is (I think)
> > > > > > 1/n^2. So the objective for a zero-info voter is to maximize
> > > > > > SUM i=1..n v_i R_i
> > > > > > with R_i = SUM i != j: 1/(n^2) (u_i - u_j).
> > > > > >
> > > > > > We also have the constraint that SUM i=1..n |v_i|^p = 1 (due to Lp
> > > > > > normalization).
> > > > > >
> > > > > > So to use a Lagrangian:
> > > > > > max SUM i=1..n R_i v_i + lambda (1 - SUM i=1..n |v_i|^p)
> > > > > > i.e.
> > > > > > max SUM i=1..n (R_i v_i - lambda |v_i|^p) + lambda
> > > > > >
> > > > > > Now do a hack and use v_i^p instead because it's easier to differentiate
> > > > > > (might not be sound?), and let's consider one particular v, say v_1.
> > > > > >
> > > > > > The derivative wrt v_1 is
> > > > > > v_1 = ( -R_1/(lambda*p) )^(1/(p-1))
> > > > > > and wrt lambda
> > > > > > sum i=1..n: v_i^p = 1.
> > > > > >
> > > > > > So what that means is that the optimum is at
> > > > > > v_i = (R_i/k)^(1/(p-1))
> > > > > > where k is a constant set so that the pth powers of the voting variables
> > > > > > sum to one. (I.e. lambda is set so that -lambda p = k, because the
> > > > > > derivative wrt lambda itself places no constraint on lambda.)
> > > > > >
> > > > > > In particular, for max norm (Range), the calculation involves an 1/infty
> > > > > > norm, i.e. 0 norm, so that the scores only depend on the sign values of
> > > > > > the R variables. I don't *quite* get the right result here (it seems to
> > > > > > indicate the optimum vote would be +1 or -1 for every candidate), which
> > > > > > I think is because I turned |v_i| into v_i above.
> > > > > >
> > > > > > For ordinary cumulative voting (l1-cumulative), all R_i are raised to
> > > > > > some power that's approaching infinity. So as this power approaches
> > > > > > infinity, the k term grows to satisfy the constraint that the pth power
> > > > > > sums must be 1. This means that everything except the v_i corresponding
> > > > > > to the greatest R_i will approach zero, whereas the remaining one
> > > > > > approaches one. So the best zero-info strategy is to give max score to
> > > > > > your favorite and nobody else.
> > > > > >
> > > > > > For the quadratic norm, v_i = R_i/k, so only here is the zero info vote
> > > > > > directly proportional to R_i.
> > > > > >
> > > > > > And R_i - R_j is proportional to u_i - u_j with the same constant of
> > > > > > proportionality throughout, because:
> > > > > > R_i - R_j = 1/(n^2) (SUM i!=k (u_i - u_k) - SUM j!=k (u_j - u_k))
> > > > > > = 1/(n^2) ( (n-1) u_i - SUM k: (u_k) + u_i - (n-1) u_j + SUM k:
> > > > > > (u_k) - u_j)
> > > > > > = 1/(n^2) (n (u_i - u_j))
> > > > > > = 1/n (u_i - u_j)
> > > > > >
> > > > > > Hence for quadratic voting, so are the optimal zero info scores v_i.
> > > > > > Looking at R_i - R_j removes the b factor, which is probably why I can't
> > > > > > show that R_i is proportional to u_i directly.
> > > > > >
> > > > > > Again, it's not entirely sound but it indicates the general direction.
> > > > > > Do improve my calculations if you can, as they're very rough.
> > > > > >
> > > > > > (The problem with quadratic voting is that it isn't cloneproof. I
> > > > > > suspect that only Range itself is, because for every other p-norm >= 1,
> > > > > > you can imagine a two-candidate election where A gets 1+epsilon points,
> > > > > > B gets 1, then clone A to make A lose, if you just make epsilon small
> > > > > > enough.)
> > > > > >
> > > > > > -km
> > > > > >
> > > ----
> > > Election-Methods mailing list - see https://electorama.com/em for list info
> >
> > --
> >
> > r b-j . _ . _ . _ . _ rbj at audioimagination.com
> >
> > "Imagination is more important than knowledge."
> >
> > .
> > .
> > .
> > ----
> > Election-Methods mailing list - see https://electorama.com/em for list info
> >
--
r b-j . _ . _ . _ . _ rbj at audioimagination.com
"Imagination is more important than knowledge."
.
.
.
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